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- Compound X (molecular formula C10H120) was treated with NH2NH2, ¯OH to yield compound Y (molecular formula C10H14). Match the 1H NMR spectra of X and Y to the corresponding structures of X and Y. Compound NH2NH2 Compound 'H NMR of X 6 H OH Y 1 H 5H 8. 6. 4 ppm or H NMR of Y 6 H 2H 5H 1 H multiplet multiplet 8. 6. 4. 3. 1 nnm 2. 2. 3, O:Compound D has the molecular formula C5H8O. Assign all peaks in 1H and 13C NMR spectrum of compound D.Identify the structures of isomers A and B (molecular formula C9H10O). Compound A: I R peak at 1742 cm−1; 1H NMR data (ppm) at 2.15 (singlet, 3 A: H), 3.70 (singlet, 2 H), and 7.20 (broad singlet, 5 H). Compound B: I R peak at 1688 cm−1; 1H NMR data (ppm) at 1.22 (triplet, 3 B: H), 2.98 (quartet, 2 H), and 7.28–7.95 (multiplet, 5 H).
- An unknown compound X has the molecular formula C6H140. Compound at 3000 cm. The 'H NMR spectral data of compound X is given below. W Here PPM values are given in delta symbol. absorption 8 ratio triplet 1.0 3 singlet 1.1 quartet singlet 1.5 3.5 3 623 6 O C) III O A) I O D) IV O B) II I 11 OH t III IVIdentify A and B, isomers of molecular formula C3H4Cl2, from the given 1H NMR data: Compound A exhibits peaks at 1.75 (doublet, 3 H, J = 6.9 Hz) and 5.89 (quartet, 1 H, J = 6.9 Hz) ppm. Compound B exhibits peaks at 4.16 (singlet, 2 H), 5.42 (doublet, 1 H, J = 1.9 Hz), and 5.59 (doublet, 1 H, J = 1.9 Hz) ppm.Identify the structures of isomers A (molecular formula C9H10O).Compound A: I R peak at 1742 cm−1; 1H NMR data (ppm) at 2.15 (singlet, 3H), 3.70 (singlet, 2 H), and 7.20 (broad singlet, 5 H).
- How could 1H NMR spectroscopy be used to distinguish among isomers A, B, and C?Treatment of alcohol A (molecular formula C5H12O) with CrO3, H2SO4, and H2O affords B with molecular formula C5H10O, which gives an IR absorption at 1718 cm−1. The 1H NMR spectrum of B contains the following signals: 1.10 (doublet, 6 H), 2.14 (singlet, 3 H), and 2.58 (septet, 1 H) ppm. What are the structures of A and B?When compound A (C5H12O) is treated with HBr, it forms compound B (C5H11Br). The 1H NMR spectrum of compound A has a 1H singlet, a 3Hdoublet, a 6H doublet, and two 1H multiplets. The 1H NMR spectrum of compound B has a 6H singlet, a 3H triplet, and a 2H quartet. Identifycompounds A and B.
- The 1H NMR spectra of two compounds, each with molecular formula C11H16, are shown here. Identify the compounds.You have a sample of a compound of molecular formula C11H15NO2, which has a benzene ring substituted by two groups, (CH3)2N – and – CO2CH2CH3, and exhibits the given 13C NMR. What disubstituted benzene isomer corresponds to these 13C data?What is the 1H NMR data (chemical shift, integration, multiplicty) of hte compound shown below?
